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(A) The coefficients of $T_r, T_{r+1}, T_{r+2}$ in $(a+b)^{12}$ are $^{12}C_{r-1}, ^{12}C_r, ^{12}C_{r+1}$.Since they are in $G.P.$,we have $(^{12}C_r

Vedclass · Accepted Answer

$283$

Vedclass · Answer

(A) The coefficients of $T_r, T_{r+1}, T_{r+2}$ in $(a+b)^{12}$ are $^{12}C_{r-1}, ^{12}C_r, ^{12}C_{r+1}$.Since they are in $G.P.$,we have $(^{12}C_r)^2 = (^{12}C_{r-1}) 	imes (^{12}C_{r+1})$.Using the property $\frac{^{n}C_k}{^{n}C_{k-1}} = \frac{n-k+1}{k}$,we get $\frac{^{12}C_r}{^{12}C_{r-1}} = \frac{12-r+1}{r} = \frac{13-r}{r}$ and $\frac{^{12}C_{r+1}}{^{12}C_r} = \frac{12-r}{r+1}$.Equating the ratios: $\frac{13-r}{r} = \frac{12-r}{r+1} \implies (13-r)(r+1) = r(12-r)$.$13r + 13 - r^2 - r = 12r - r^2 \implies 12r + 13 = 12r \implies 13 = 0$,which is impossible.Thus,there are no such values of $r$,so $p = 0$.For the expansion $(3^{1/4} + 4^{1/3})^{12}$,the general term is $T_{k+1} = ^{12}C_k (3^{1/4})^{12-k} (4^{1/3})^k = ^{12}C_k \cdot 3^{(12-k)/4} \cdot 4^{k/3}$.For the term to be rational,$(12-k)$ must be divisible by $4$ and $k$ must be divisible by $3$.Possible values for $k \in \{0, 1, 2, \dots, 12\}$ are $k=0$ and $k=12$.For $k=0$: $T_1 = ^{12}C_0 \cdot 3^3 \cdot 4^0 = 1 \cdot 27 \cdot 1 = 27$.For $k=12$: $T_{13} = ^{12}C_{12} \cdot 3^0 \cdot 4^4 = 1 \cdot 1 \cdot 256 = 256$.Sum $q = 27 + 256 = 283$.Therefore,$p+q = 0 + 283 = 283$.

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